1.51 problem 51

Internal problem ID [8015]
Book : First order enumerated odes
Section : section 1
Problem number : 51
Date solved : Monday, October 21, 2024 at 04:40:55 PM
CAS classification : [[_homogeneous, `class A`], _rational, _dAlembert]

Solve

\begin{align*} {y^{\prime }}^{2}&=\frac {y}{x} \end{align*}

1.51.1 Solved as first order homogeneous class A ode

Time used: 0.763 (sec)

Solving for \(y^{\prime }\) gives

\begin{align*} y^{\prime }&=\frac {\sqrt {x y}}{x}\tag {1} \\ y^{\prime }&=-\frac {\sqrt {x y}}{x}\tag {2} \end{align*}

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {\sqrt {x y}}{x}\tag {1} \end{align*}

An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n y)= t^n f(x,y) \]

In this case, it can be seen that both \(M=\sqrt {x y}\) and \(N=x\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence

\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(y=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= \sqrt {u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {\sqrt {u \left (x \right )}-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {\sqrt {u \left (x \right )}-u \left (x \right )}{x} = 0 \]

Or

\[ u^{\prime }\left (x \right ) x -\sqrt {u \left (x \right )}+u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode \(u^{\prime }\left (x \right ) = \frac {\sqrt {u \left (x \right )}-u \left (x \right )}{x}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= \frac {\sqrt {u \left (x \right )}-u \left (x \right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= \sqrt {u}-u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{\sqrt {u}-u}\,du} &= \int { \frac {1}{x} \,dx}\\ -2 \ln \left (\sqrt {u \left (x \right )}-1\right )&=\ln \left (x \right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(\sqrt {u}-u=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0\\ u \left (x \right )&=1 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -2 \ln \left (\sqrt {u \left (x \right )}-1\right ) = \ln \left (x \right )+c_1\\ u \left (x \right ) = 0\\ u \left (x \right ) = 1 \end{align*}

Converting \(-2 \ln \left (\sqrt {u \left (x \right )}-1\right ) = \ln \left (x \right )+c_1\) back to \(y\) gives

\begin{align*} -2 \ln \left (\sqrt {\frac {y}{x}}-1\right ) = \ln \left (x \right )+c_1 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Converting \(u \left (x \right ) = 1\) back to \(y\) gives

\begin{align*} y = x \end{align*}

In canonical form, the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {\sqrt {x y}}{x}\tag {1} \end{align*}

An ode of the form \(y' = \frac {M(x,y)}{N(x,y)}\) is called homogeneous if the functions \(M(x,y)\) and \(N(x,y)\) are both homogeneous functions and of the same order. Recall that a function \(f(x,y)\) is homogeneous of order \(n\) if

\[ f(t^n x, t^n y)= t^n f(x,y) \]

In this case, it can be seen that both \(M=-\sqrt {x y}\) and \(N=x\) are both homogeneous and of the same order \(n=1\). Therefore this is a homogeneous ode. Since this ode is homogeneous, it is converted to separable ODE using the substitution \(u=\frac {y}{x}\), or \(y=ux\). Hence

\[ \frac { \mathop {\mathrm {d}y}}{\mathop {\mathrm {d}x}}= \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u \]

Applying the transformation \(y=ux\) to the above ODE in (1) gives

\begin{align*} \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}}x + u &= -\sqrt {u}\\ \frac { \mathop {\mathrm {d}u}}{\mathop {\mathrm {d}x}} &= \frac {-\sqrt {u \left (x \right )}-u \left (x \right )}{x} \end{align*}

Or

\[ u^{\prime }\left (x \right )-\frac {-\sqrt {u \left (x \right )}-u \left (x \right )}{x} = 0 \]

Or

\[ u^{\prime }\left (x \right ) x +\sqrt {u \left (x \right )}+u \left (x \right ) = 0 \]

Which is now solved as separable in \(u \left (x \right )\).

The ode \(u^{\prime }\left (x \right ) = -\frac {\sqrt {u \left (x \right )}+u \left (x \right )}{x}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {\sqrt {u \left (x \right )}+u \left (x \right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= \frac {1}{x}\\ g(u) &= -\sqrt {u}-u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{-\sqrt {u}-u}\,du} &= \int { \frac {1}{x} \,dx}\\ \ln \left (\frac {1}{\left (\sqrt {u \left (x \right )}+1\right )^{2}}\right )&=\ln \left (x \right )+c_2 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(-\sqrt {u}-u=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (\frac {1}{\left (\sqrt {u \left (x \right )}+1\right )^{2}}\right ) = \ln \left (x \right )+c_2\\ u \left (x \right ) = 0 \end{align*}

Converting \(\ln \left (\frac {1}{\left (\sqrt {u \left (x \right )}+1\right )^{2}}\right ) = \ln \left (x \right )+c_2\) back to \(y\) gives

\begin{align*} \ln \left (\frac {1}{\left (\sqrt {\frac {y}{x}}+1\right )^{2}}\right ) = \ln \left (x \right )+c_2 \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=0\\ y&=x\\ y&=\left (\frac {2 x \,{\mathrm e}^{c_1} \left (\sqrt {x \,{\mathrm e}^{c_1}}-1\right )}{\sqrt {x \,{\mathrm e}^{c_1}}}-x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1}\\ y&=\left (\frac {2 x \,{\mathrm e}^{c_1} \left (\sqrt {x \,{\mathrm e}^{c_1}}+1\right )}{\sqrt {x \,{\mathrm e}^{c_1}}}-x \,{\mathrm e}^{c_1}+1\right ) {\mathrm e}^{-c_1}\\ y&=-\left (-\frac {2 x \,{\mathrm e}^{c_2} \left (\sqrt {x \,{\mathrm e}^{c_2}}-1\right )}{\sqrt {x \,{\mathrm e}^{c_2}}}+x \,{\mathrm e}^{c_2}-1\right ) {\mathrm e}^{-c_2}\\ y&=-\left (-\frac {2 x \,{\mathrm e}^{c_2} \left (\sqrt {x \,{\mathrm e}^{c_2}}+1\right )}{\sqrt {x \,{\mathrm e}^{c_2}}}+x \,{\mathrm e}^{c_2}-1\right ) {\mathrm e}^{-c_2} \end{align*}

1.51.2 Solved as first order ode of type nonlinear p but separable

Time used: 0.181 (sec)

The ode has the form

\begin{align*} (y')^{\frac {n}{m}} &= f(x) g(y)\tag {1} \end{align*}

Where \(n=2, m=1, f=\frac {1}{x} , g=y\). Hence the ode is

\begin{align*} (y')^{2} &= \frac {y}{x} \end{align*}

Solving for \(y^{\prime }\) from (1) gives

\begin{align*} y^{\prime } &=\sqrt {f g}\\ y^{\prime } &=-\sqrt {f g} \end{align*}

To be able to solve as separable ode, we have to now assume that \(f>0,g>0\).

\begin{align*} \frac {1}{x} &> 0\\ y &> 0 \end{align*}

Under the above assumption the differential equations become separable and can be written as

\begin{align*} y^{\prime } &=\sqrt {f}\, \sqrt {g}\\ y^{\prime } &=-\sqrt {f}\, \sqrt {g} \end{align*}

Therefore

\begin{align*} \frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx\\ -\frac {1}{\sqrt {g}} \, dy &= \left (\sqrt {f}\right )\,dx \end{align*}

Replacing \(f(x),g(y)\) by their values gives

\begin{align*} \frac {1}{\sqrt {y}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx\\ -\frac {1}{\sqrt {y}} \, dy &= \left (\sqrt {\frac {1}{x}}\right )\,dx \end{align*}

Integrating now gives the following solutions

\begin{align*} \int \frac {1}{\sqrt {y}}d y &= \int \sqrt {\frac {1}{x}}d x +c_1\\ 2 \sqrt {y} &= 2 x \sqrt {\frac {1}{x}}\\ \int -\frac {1}{\sqrt {y}}d y &= \int \sqrt {\frac {1}{x}}d x +c_1\\ -2 \sqrt {y} &= 2 x \sqrt {\frac {1}{x}} \end{align*}

Therefore

\begin{align*} y &= x \sqrt {\frac {1}{x}}\, c_1 +\frac {c_1^{2}}{4}+x \\ y &= x \sqrt {\frac {1}{x}}\, c_1 +\frac {c_1^{2}}{4}+x \\ \end{align*}

1.51.3 Solved as first order ode of type dAlembert

Time used: 0.136 (sec)

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} p^{2} = \frac {y}{x} \end{align*}

Solving for \(y\) from the above results in

\begin{align*} y &= p^{2} x\tag {1A} \end{align*}

This has the form

\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= p^{2}\\ g &= 0 \end{align*}

Hence (2) becomes

\begin{align*} -p^{2}+p = 2 x p p^{\prime }\left (x \right )\tag {2A} \end{align*}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

\begin{align*} -p^{2}+p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0\\ p_{2} &=1 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

\begin{align*} y = 0\\ y = x \end{align*}

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{align*} p^{\prime }\left (x \right ) = \frac {-p \left (x \right )^{2}+p \left (x \right )}{2 x p \left (x \right )}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. In canonical form a linear first order is

\begin{align*} p^{\prime }\left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {1}{2 x}\\ p(x) &=\frac {1}{2 x} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{2 x}d x}\\ &= \sqrt {x} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \left (\mu \right ) \left (\frac {1}{2 x}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (p \sqrt {x}\right ) &= \left (\sqrt {x}\right ) \left (\frac {1}{2 x}\right ) \\ \mathrm {d} \left (p \sqrt {x}\right ) &= \left (\frac {1}{2 \sqrt {x}}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} p \sqrt {x}&= \int {\frac {1}{2 \sqrt {x}} \,dx} \\ &=\sqrt {x} + c_1 \end{align*}

Dividing throughout by the integrating factor \(\sqrt {x}\) gives the final solution

\[ p \left (x \right ) = \frac {\sqrt {x}+c_1}{\sqrt {x}} \]

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y = \left (\sqrt {x}+c_1 \right )^{2}\\ \end{align*}

1.51.4 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & {y^{\prime }}^{2}=\frac {y}{x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=\frac {\sqrt {x y}}{x}, y^{\prime }=-\frac {\sqrt {x y}}{x}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\sqrt {x y}}{x} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\sqrt {x y}}{x} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

1.51.5 Maple trace

Methods for first order ODEs:
 

1.51.6 Maple dsolve solution

Solving time : 0.022 (sec)
Leaf size : 39

dsolve(diff(y(x),x)^2 = y(x)/x, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= \frac {\left (x +\sqrt {c_1 x}\right )^{2}}{x} \\ y &= \frac {\left (-x +\sqrt {c_1 x}\right )^{2}}{x} \\ \end{align*}

1.51.7 Mathematica DSolve solution

Solving time : 0.047 (sec)
Leaf size : 46

DSolve[{(D[y[x],x])^2==y[x]/x,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \frac {1}{4} \left (-2 \sqrt {x}+c_1\right ){}^2 \\ y(x)\to \frac {1}{4} \left (2 \sqrt {x}+c_1\right ){}^2 \\ y(x)\to 0 \\ \end{align*}